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How Good is the Leighton-Rao Relaxation?

Continuing our seemingly never-ending series on approximations to edge expansion, we come to the question of how good is the Leighton-Rao relaxation.

Given a graph , we are trying to approximate the sparsest cut problem (which, in turn, approximates the edge expansion problem within a factor of two) defined as

And we have seen that an equivalent characterization of sparsest cut is:

In the Leighton-Rao relaxation, we have

where is the set of all semi-metrics, that is, of all functions such that , , and that obey the “triangle inequality”

for all .

This is clearly a relaxation of the sparsest cut problem, because for every the “distances” define indeed a semi-metric over .

How bad can the relaxation be? Take a family of regular constant-degree expander graphs, that is graphs where is a constant and the degree is a constant independent of . Define to be the shortest path distance between and in the graph. This is clearly a semi-metric, and we have

because along every edge the distance is precisely one, and

because, for every vertex , at most other vertices can be within
distance , and so at least half the vertices are at distance .

So we have

even though , so the error in the approximation can be as bad as a factor of . We shall prove that this is as bad as it gets.

Our task is then: given a feasible solution to (2), that is, a distance function defined on the vertices of , find a feasible solution to sparsest cut whose cost is as small as possible, and no more than a factor off from the cost of the solution to (2) that we started from. Instead of directly looking for a cut, we shall look for a solution to (1), which is a more flexible formulation.

The general strategy will be, given , to find a distribution over vectors such that for every two vertices we have

Suppose that we start from a solution to (2) of cost , that is, we have a semimetric such that

Then our distribution is such that

and

so

and there must exist a vector in the support of the distribution such that

and now we are done because the cost of in (1) is at most times the optimum of the Leighton-Rao relaxation.

Though this works, it seems an overkill to require condition (3) for all pairs of vertices. It seems sufficient to require

only for edges, and to require

only “on average” over all pairs of vertices. It can be shown, however, that if one wants to round a generalization of the Leighton-Rao relaxation to sparsest cut with “non-uniform demands,” then any rounding algorithm can be turned into a rounding algorithm that satisfies (3).

So how do we start from an arbitrary metric and come up with a probabilistic linear order of the vertices so that their distances in the original metric are well approximated in the linear ordering? We will describe a method of Bourgain, whose applicability in this setting is due Linial, London and Rabinovich.

The starting point is to see that if is a set of vertices, and we define

then, no matter how we choose , we always have

[Hint: use the triangle inequality to see that .]

This means that “all we have to do” is to find a distribution over sets such that for every we have

If you want to make it sound more high-brow, you may call such a distribution a Frechet embedding of into . Constructing such a distribution will be the subject of the next post.